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HDOJ1002 A + B Problem II

By Wu Yudong on May 08, 2018

Problem - 1002

Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899   998877665544332211
Sample Output
Case 1: 1 + 2 = 3
Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

下面是我的AC代码:

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>

char a[1001],b[1001];

void sum(char *s1, char *s2) //加法运算函数
{
    int len1,len2,i,t,flag;
    char *p,*q;
    int c[1003]={0};
    len1=strlen(s1); len2=strlen(s2);
	//使用字符串存储加数与被加数
    i=t=flag=0;
    printf("%s + ",a); //控制输出语句
    printf("%s = ",b);
    p=strrev(s1); //转置字符串
    q=strrev(s2);
    while(*p || *q){
        if(*p) t+=*p-'0';
        if(*q) t+=*q-'0';
        t+=flag;
        if(t>=10){t=t%10; c[i++]=t; flag=1;}
        else {c[i++]=t; flag=0;}
        p++; q++;
        t=0;
    }
    if(flag) c[i]+=1;
    else i--;

    for(; i>=0; i--){
        printf("%d",c[i]); //输出结果
    }
}

void solve()
{
    int i,n,t;
    i=1;
    scanf("%d",&n);
    t=n;
    getchar();
    while(n--){
        scanf("%s",a);  scanf("%s",b);
        printf("Case %d:\n",i);
        sum(a,b);
        if(i!=t) printf("\n\n");
        else printf("\n");
        i++;
        memset(a,'\0',strlen(a)*sizeof(char)); //清空字符串
        memset(b,'\0',strlen(b)*sizeof(char));
    }
}

int main()
{
    solve();
    getchar();
    getchar();
    return 0;
}

 

 

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